获取指定节点类型的最新10篇文章？

Question

现在,如果节点类型是文章.在术语页面左侧,我想显示最新的10篇文章的标题,哪个节点类型是文章.我不想使用意见,我该怎么办？谢谢.

如果我想在节点页面的左边显示最新的10篇文章的标题,哪个节点类型是文章.如何编写查询.非常感谢.

ps:我发现EntityFieldQuery也许可以做到这一点,但我现在不知道如何做到这一点.

我的代码:

$query = new EntityFieldQuery();

$query
 ->entityCondition('entity_type', 'node')
 ->entityCondition('bundle', 'article')
 ->propertyCondition('status', 1)
 ->propertyOrderBy('created', 'DESC')
  ->range(0, 10);

$result = $query->execute();

Answer 1

代码可以是这样的(使用db_select())

$query = db_select("node", "n") // select from the node table
    ->fields("n", array("nid", "title")) // fields nid, title
    ->condition("type", "page", "=") // where the node type = page
    ->orderBy("created", "DESC") // order by the newest
    ->range(0, 10) // select only 10 records
    ->execute(); // execute the query

while($record = $query->fetchAssoc()) {
    print(l($record['title'], "node/" . $record['nid'])); // print the node title linked to node.
}

使用EntityFieldQuery()的另一个例子:

$query = new EntityFieldQuery();
$entities = $query->entityCondition('entity_type', 'node')          
      ->entityCondition('bundle', 'club')
      ->propertyOrderBy("created", "DESC")
      ->range(0, 10)
      ->execute();

foreach($entities['node'] as $obj)
{
    $node = node_load($obj->nid);
    print(l($node->title, "node/" . $node->nid));
}

性能明智:使用第一种方法.

希望这有助于......穆罕默德.